vlcek@mit-caf.MIT.EDU (Jim Vlcek) (08/03/88)
Hal Pomeranz was asking about array references and pointers to arrays,
namely, why doesn't
char array[10], **ptr;
ptr = &array;
work? The reason is that array names are not variables, they are
compile time constants. Hence, taking the address of an array name
is a meaningless operation, since there is no allocated storage to
take the address of. The proper way to think of an array name is as a
pointer-valued *constant*, not as an actual pointer.
Read page 94 of the old K&R, I don't know where it is in the new one,
for a discussion of arrays vs. pointers.
Oh, one more thing, in case you're not yet confused. If you declare
an ``array'' in an argument list, you *do* in fact get a pointer, and
you can manipulate it as such. For example,
foo (s) char s[];
{
while (*s++ != '\0')
{
...
}
}
is quite correct. This is because arrays are passed to functions as
pointers, and, due to the call-by-value nature of C function calls,
the parameter can be modified safely within the body of the called
function. (That such a construct could not be a constant should also
be immediately clear, since the value of ``s'' above could not be
known at compile time, as it would for an actual array.)
--
Jim Vlcek
vlcek@caf.mit.edu
!{ihnp4,harvard,seismo,rutgers}!mit-eddie!mit-caf!vlcek